Question

The position of a partical along X- axis at time t is given by x=2+t—3t² . The displacement and the the distance travelled in the interval , t = 0 to t = 1 are respectively
(a) 2 , 2
(b) –2 , 2.5
(c) 0 , 2
(d) –2 , 2.1

Please Answer And Explain It ...​

Answer 1

Answer:

Explanation:

x = 2 +t -3t² ---------(1)

differentiate wrt time

dx/dt = 1 -6t ----------(2)

again differentiate wrt time

d²x/dt² = -6

hence, motion is retarding becoz d²x/dt² < 0

let at t time velocity of particle is zero

e.g dx/dt = 0 = 1-6t = 1/6

at t = 1/6 velocity of particle is zero .

then particle moves in back direction at t = 1 sec

now,

displacement = final position - initial position = x(1) - x (0)

= 2 + 1 -3 - 2

= -2 m

distance = | x(1) - x (1/6) | + | x (1/6) - x(0) |

= | 2 + 1 -3 -(2 +1/6 -3/36 | + | 2 + 1/6 -3/36 - 2 |

=| -2 - 1/6 +1/12 | + | 2 +1/6 -1/12 -2 |

= | -2 -1/12 | + | 1 /12 |

= 2 + 1/12 + 1/12

= 2 + 1/6 = 13/6 m